Bài 1: 

b: \(\Leftrightarrow x-2=0\)

hay x=2

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

\(y^2+4^x+2y-2^{x+1}+2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(4^x-2^{x+1}+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}\)

\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)

\(\Leftrightarrow\frac{x^2+4x+4+2}{x+2}+\frac{x^2+16x+64+8}{x+8}=\frac{x^2+8x+16+4}{x+4}+\frac{x^2+12x+36+6}{x+6}\)

\(\Leftrightarrow2x+10+\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)

\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)

Tới đây quy đồng làm tiếp nhé

=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)

=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)

=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)

Voi \(\frac{1}{x+2}-....\)=0 ta co

Dat x+5=t

=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0

=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)

=>t=0

=>x=-5

Vay phuong trinh co nghiem x=0;-5

toán lớp 8 mà đi giải phương trình hả má

ĐKXĐ:\(x\ne-2;-4;-6;-8\)

\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)

\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)

\(\frac{10+32}{\left(x+2\right)\left(x+8\right)}=\frac{10x+48}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\frac{5x+16}{\left(x+2\right)\left(x+8\right)}=\frac{5x+24}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\left(5x+16\right)\left(x+4\right)\left(x+6\right)=\left(5x+24\right)\left(x+2\right)\left(x+8\right)\)

\(\Leftrightarrow x^2+5x=0\)(bạn tự biến đổi nhé)

\(\Leftrightarrow x=0;-5\)(tm ĐKXĐ)

Vậy phương trình có nghiệm 0;-5 (mình làm hơi tắt bn thông cảm nha)

=> \(\frac{(x+2)^2+2}{x+2}+\frac{(x+8)^2+8}{x+8}=\frac{(x+4)+4}{x+4}+\frac{(x+6)^2+6}{x+6}\)

=> 2x + 10 + \(\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)

=>-x \((\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8})=0\)

                              \(x=0\)

\(=>\orbr{\frac{1}{x+2}}-.....+\frac{1}{x+8}=0\)

Với \(\frac{1}{x+2}-...=0\). Ta có :

Đặt x + 5 = t

=> \(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}=0\)

\(=>2t(\frac{1}{t^2-1}+\frac{1}{t^2-9})=0\)

=> t = 0

=> x = -5

Vậy phương trình có nghiệm x= 0 ; - 5

\(\Leftrightarrow\sqrt{x^2\left(x^2-8x+16\right)}+\left|\left(x-4\right)^2\right|=4\\ \)

\(\Leftrightarrow\left|x\right|\sqrt{\left(x-4\right)^2}+\left(x-4\right)^2=4\)

\(\Leftrightarrow\left|x\left(x-4\right)\right|+x^2-8x+16=4\)(1)

KL: PT có 2 nghiệm là x = 3 và x = \(3+\sqrt{3}\).